10. The velocity of an angry badger is given by€v = 2t 2 + 4t − 3 . What is the badger’s positionand acceleration at 3 seconds?

To find the badger's position and acceleration at 3 seconds, we need to follow these steps:

Step 1: Find the badger's position function by integrating the velocity function. The position function is given by:

s = ∫v dt

Step 2: Differentiate the velocity function to find the acceleration function. The acceleration function is given by:

a = dv/dt

Step 3: Substitute the value of t = 3 into the position and acceleration functions to find the badger's position and acceleration at 3 seconds.

Let's calculate each step:

Step 1: Integrating the velocity function: To integrate 2t^2 + 4t - 3, we use the power rule of integration. The integral of t^n is (1/n+1) * t^(n+1).

∫v dt = ∫(2t^2 + 4t - 3) dt = (2/3)t^3 + 2t^2 - 3t + C

So, the position function is s = (2/3)t^3 + 2t^2 - 3t + C.

Step 2: Differentiating the velocity function: To differentiate 2t^2 + 4t - 3, we use the power rule of differentiation. The derivative of t^n is n * t^(n-1).

a = d/dt (2t^2 + 4t - 3) = 4t + 4

So, the acceleration function is a = 4t + 4.

Step 3: Substitute t = 3 into the position and acceleration functions: To find the badger's position and acceleration at 3 seconds, we substitute t = 3 into the position and acceleration functions we found in steps 1 and 2.

Position at 3 seconds: s(3) = (2/3)(3)^3 + 2(3)^2 - 3(3) + C = 18 + 18 - 9 + C = 27 + C

Acceleration at 3 seconds: a(3) = 4(3) + 4 = 12 + 4 = 16

So, the badger's position at 3 seconds is 27 + C and the acceleration is 16.